Question

Let the function $f:R\to R$ satisfies $f\left(x\right)={\int }_0^x(\frac{1}{e^{-t}}+f(x-t\left)\right)dt$.
[Note : e denotes Napier's constant]
Which of the following statement(s) is/are correct?

• A. Number of solutions of the equation $f\left(x\right)+x^2=0$ is two
• B. The value of ${\int }_0^{1}f\left(x\right)dx$ is equal to 1.
• C. $f\left(x\right)$ is an odd function
• D. The equation $f\left(x\right)-f^{\prime }\left(x\right)=0$ has no real solution.

Answer 1

The correct option is A Number of solutions of the equation $f\left(x\right)+x^2=0$ is two

Given : $f:R\to Rf\left(x\right)={\int }_0^x[\frac{1}{e^{-t}}+f(x-t\left)\right]dt$

$f\left(x\right)={\int }_0^x[\frac{1}{e^{-t}}+f(x-t\left)\right]dtf\left(x\right)={\int }_0^x{e}^{t}dt-{\int }_0^{x}f(x-t)dtf\left(x\right)={\int }_0^x{e}^{t}dt+{\int }_0^{x}f\left(t\right)dt$

Using leibntz rule

$f^{\prime }\left(x\right)=e^x+f\left(x\right)f^{\prime }\left(x\right)-f\left(x\right)=e^x$

Multiplying both sides with $e^{-x}$

$[-f(x)+f^{\prime }(x\left)\right]e^x=1$

Integrating with respect to $x$

$\int [-f(x)+f^{\prime }(x\left)\right]e^{x}dx=\int dx{e}^{-x}f\left(x\right)=xf\left(x\right)=x{e}^x$

$\left(A\right)f\left(x\right)+x^2=x{e}^x+x^2$ Are solution

$\left(A\right)f\left(x\right)+x^2=x{e}^x+x^2\left(B\right){\int }_0^{1}f\left(x\right)dx={\int }_0^{1}x{e}^{x}dx={[{xe}^x-e^x]}_0^1=0-(0-e^0)=1\left(C\right)f\left(x\right)=x{e}^{x}f(-x)-x{e}^{-x}\left(Notodd\right)\left(D\right)f\left(x\right)-f^{\prime }\left(x\right)=x{e}^{-x}-x{e}^{-x}-e^x-e^x=0e^x=0x=-\infty \left(Norealsolution\right)$